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The Baire category theorem states that a complete metric space is a Baire space: the intersection of countably many open dense sets is dense. Equivalently, a nonempty open subset cannot be a countable union of nowhere dense sets.

Let $U_n$ be open dense and $O$ any nonempty open set. Choose a closed ball $B_1$ of positive radius at most $1/2$ inside $O\cap U_1$. Inductively choose $B_{n+1}$ inside the interior of $B_n$ intersected with $U_{n+1}$, with positive radius at most $2^{-(n+1)}$. Density makes this intersection nonempty, and openness supplies a sufficiently small closed ball. The centres form a Cauchy sequence because all later centres lie in $B_n$, whose diameter tends to zero. Completeness gives a limit $x$. Since each ball is closed and contains all later centres, $x\in B_n$ for every $n$. Thus $x\in O\cap {\bigcap }_n{U}_n$, proving density. Applying this to complements of the closures of nowhere dense sets gives the equivalent formulation.

Write $f=X^2+bX+c$ and $g=X^2+BX+C$. Since their common square-free discriminant $d$ is congruent to zero or one modulo four, it must be one modulo four; thus $b,B$ are odd. It is not a rational square because a square-free integer square is only zero or one. Both polynomials are irreducible over $\mathbb{Q}$, and their roots generate the same quadratic extension $\mathbb{Q}(\sqrt{d})$. Their integral roots are $(-b+\sqrt{d})/2$ and $(-B+\sqrt{d})/2$, respectively. Each generated order isthe usual ring of integers for square-free $d\equiv 1(\mathrm{mod}4)$. This identifies both quotients with the same ring. Explicitly, if $\xi$ denotes the image of $X$ modulo $g$, send the image of $X$ modulo $f$ to $\xi +(B-b)/2$. Equal discriminants give $f(\xi +(B-b)/2)=g(\xi )=0$, and the opposite translation gives its inverse. The two quotient rings are isomorphic.

The cubic has no rational root: its only possible integer roots are $\pm 1$, and neither works. It is therefore irreducible. Let $\alpha$ be its image in $K$; it is an algebraic integer, so $\mathbb{Z}[\alpha ]\subseteq {\mathcal{O}}_K$. The power-basis discriminant isThe index-discriminant formula from part (i) says that the square of $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$ divides 59. It must be one. Hence $\boxed{{\mathcal{O}}_K=\mathbb{Z}[\alpha ]}$, with integral basis $1,\alpha ,{\alpha }^2$ and field discriminant $-59$.

For a basis of a module $m_1,\dots ,m_d$ of the full-rank module $M$, define its discriminant of elements of a number field byThe field discriminant is this determinant for an integral basis of ${\mathcal{O}}_K$. A change of integral basis with determinant $\pm 1$ does not change it. Write $m_j={\sum }_i{b}_{ij}{e}_i$ in an integral basis and let $B=(b_{ij})$. The two trace matrices satisfy $T_M=B^{\mathsf{T}}{T}_{{\mathcal{O}}_K}B$. The index of the lattice is $r=|\det B|$, as follows for example from Smith normal form. ThereforeMultiplicative closure makes $M$ an order when it contains one, but the determinant argument itself requires only full finite index, so it also covers the stated module hypothesis without that extra condition.

Take a maximal matching in a graph with $m$ edges and let $S$ be its $2m$ endpoints. Maximality makes its complement an independent set, since an edge wholly outside $S$ could be added. Every vertex outside $S$ therefore sends all its $k$ edges into $S$, giving $k(n-2m)$ crossing edges. Every vertex in $S$ already uses at least one incident edge inside $S$, its matched edge, so it sends at most $k-1$ outside. HenceThe maximum matching in a graph is at least this large. For $k=2$, take any disjoint union of $r$ triangles. It is 2-regular, has $n=3r$ vertices and matching number $r=n/3$, since each triangle contributes exactly one edge. As $r$ varies this gives infinitely many equality examples.

Let $A,B$ be the two parts of the bipartite graph. Counting edges gives $k|A|=k|B|$, hence $|A|=|B|=n/2$. For $S\subseteq A$, the $k|S|$ edges incident to $S$ all end in $N(S)$, whose vertices have total capacity $k|N(S)|$. Thus $|N(S)|\ge |S|$. Hall's marriage theorem gives a perfect matching, and no matching in a graph can have more than $n/2$ edges. The matching number is $\boxed{n/2}$.

Hall's marriage theorem states that a finite bipartite graph with parts $A,B$ has a matching in a graph saturating $A$ exactly when $|N(S)|\ge |S|$ for every $S\subseteq A$. Necessity follows because the matched neighbours of $S$ are distinct.

For sufficiency use induction on $|A|$. The empty case is immediate. If a nonempty proper subset $S$ has $|N(S)|=|S|$, the induced graph on $S,N(S)$ satisfies Hall's condition and has a saturating matching in a graph by induction. The remaining graph on $A\setminus S,B\setminus N(S)$ also satisfies it: for $T\subseteq A\setminus S$,Apply induction again and unite the two matchings in a graph. If there is no such tight proper subset, every nonempty proper $S$ has $|N(S)|\ge |S|+1$. Choose a vertex $a$ and a neighbour $b$, which exists by Hall's condition. Delete both. Every remaining subset loses at most one neighbour and still satisfies Hall's condition. Induction supplies a matching in a graph of the remainder; add $ab$. For $|A|=1$ this last step directly proves the assertion.

For constant equation-of-state parameter $w$, the cosmological continuity equation integrates to $\rho ={\rho }_0{a}^{-3(1+w)}$ using today's normalization. In conformal time, $\mathcal{H}=a^{\prime }/a=\dot{a}$, so the Friedmann equation becomesSince ${\mathcal{H}}^{\prime }=a\ddot{a}$, the Friedmann acceleration equation givesFor $k{c}^2=1$, put $q=(1+3w)/2$. When $q\ne 0$, integration gives $\mathcal{H}=\cot (q(\tau -{\tau }_0))$. Integrating $a^{\prime }/a=\mathcal{H}$ then yields the closed constant-equation-of-state Friedmann scale factorChoose a connected interval with positive sine; the origin of conformal time can absorb ${\tau }_0$. For $w>-1/3$ this is the expansion and recollapse branch between consecutive zeros of the sine. Negative $q$ instead gives the corresponding minimum-radius branch on its positive-sine interval. The printed power formula excludes $w=-1/3$: in that exceptional case $\mathcal{H}=H_0$ is constant and $a=A{e}^{H_0\tau }$, with $H_0^2+1=8\pi G{\rho }_0/3$. A real solution requires the latter constant to be at least one; equality permits a static solution.

For pressure-free matter, $w=0$, so $\beta =1/2$. Set the big-bang origins to zero. The closed matter-dominated Friedmann solution follows by ${\sin }^2(\beta \tau )=(1-\cos 2\beta \tau )/2$ and integrating $dt=ad\tau$:The entire cycle is $0\le 2\beta \tau \le 2\pi$; maximum expansion has $a=\alpha$ at $2\beta \tau =\pi$, and the big crunch occurs at $t=\alpha \pi /(2\beta )$.

Multiply the Friedmann equation by $a^2$ and differentiate:Insert the cosmological continuity equation to obtainOn a nonstatic branch divide by $2a\dot{a}$; continuity extends the result across isolated turning points. The resulting Friedmann acceleration equation isAn everywhere static branch needs the independent spatial Einstein field equations: the two differentiated constraints alone cannot determine the pressure when $\dot{a}\equiv 0$. This is the qualification to deriving the acceleration equation solely by division.

By Gibbs inequality,with equality exactly when $\boxed{a_j=p_j}$. Put $q_j=u_j/U$, a probability distribution. The optimal log-optimal multiplicative betting rate isFor $U<1$ this is strictly positive by Gibbs inequality; taking $0<\epsilon <g_{\ast }$ in the preceding strong law of large numbers bound proves growth with probability tending to one.

For $U>1$ the casino can set $u_j=U{p}_j$. Every optimal return multiplier is then exactly $1/U$, so the fortune decreases deterministically as $f{U}^{-m}$. More generally $g_{\ast }=-\log U<0$ for that choice. An arbitrary choice with $U>1$ need not cause decline. For example, take $p_1=p_2=1/2$, $u_1=0.01$, $u_2=1.99$. Then $U=2$ but $g_{\ast }=(1/2)\log [0.25/(0.01\cdot 1.99)]>0$. For any fixed positive $p$, making one $u_j$ sufficiently small while preserving $U$ also makes the optimal rate positive.

Let $J_1,J_2,\dots$ be independent rolls with the specified probability distribution, and let $F_m$ be the fortune after $m$ games. Its multiplicative evolution givesby the strong law of large numbers. The summands have finitely many finite values, so its hypotheses hold. For any $\epsilon >0$, the probability that $f{e}^{m(g-\epsilon )}\le F_m\le f{e}^{m(g+\epsilon )}$ tends to one. Thus maximizing $g(a)$ maximizes the typical exponential growth rate, the objective of log-optimal multiplicative betting. Maximizing expected fortune would be a different objective and would not control the typical repeated-game outcome.

For probability distributions $p,q$ with positive entries, Gibbs inequality iswith equality exactly when $p=q$. Indeed, $\log x\le x-1$, with equality only at $x=1$, givesThis also proves the equality condition. With zero entries the same Kullback-Leibler divergence statement holds using $0\log (0/q)=0$ and $p\log (p/0)=+\infty$ for $p>0$, or by restricting to the support and taking limits.

The applied pressure changes the dynamic condition to ${\varphi }_t+g\eta =-p/\rho$. For its specified spatial mode, the two amplitude equations becomeWhen $\Omega \ne \omega$, the corresponding particular oscillation isFree oscillations with amplitudes $C\cos \omega t+D\sin \omega t$ may be added. If the mode starts from rest at $t=0$, its total amplitude instead isNear resonance, the response grows too large for the linearization. At exact resonance the rest-start limit is $A=(\epsilon h/2)(\omega t\cos \omega t-\sin \omega t)$, with secular growth. Dissipation or nonlinear effects would be required to limit it.

For small irrotational surface gravity waves, write the velocity field as $\nabla \varphi$. Incompressibility gives ${\nabla }^2\varphi =0$. The rigid walls have ${\varphi }_x=0$ at $x=0,a$, ${\varphi }_y=0$ at $y=0,b$, and ${\varphi }_z=0$ at $z=-h$. At $z=0$, the linear kinematic boundary condition and pressure condition areSeparate variables for a rectangular standing surface-gravity mode:The horizontal cosine factors and vertical hyperbolic cosine satisfy all rigid-wall conditions. Put $\eta =A(t)\cos (m\pi x/a)\cos (n\pi y/b)$ and $\lambda =k\tanh kh$. The free-surface conditions give $\dot{A}=\lambda B$ and $\dot{B}=-gA$, henceBoth integers are nonnegative and not both zero. The zero pair would be a spatially uniform displacement, excluded by fixed fluid volume; a spatially constant potential is only a gauge mode.

Take the circle with the usual positive, counterclockwise orientation. The only singularity is at zero, and the Laurent series givesThe coefficient of $z^{-1}$ is $1/2$, so the residue theorem givesThe essential singularity at zero still has a well-defined residue; the theorem does not require it to be a pole.

False. The rational numbers $\mathbb{Q}$ with their subspace topology from $\mathbb{R}$ are Hausdorff. If a subset contains two distinct rationals $a<b$, choose an irrational $c$ between them; intersecting the subset with $(-\infty ,c)$ and $(c,\infty )$ gives a separation into two nonempty relatively open sets. Thus every nonempty connected subset of $\mathbb{Q}$ is a singleton, so it is totally disconnected. Yet no rational is isolated, because every interval around it contains another rational. Therefore $\mathbb{Q}$ is not a discrete space. Hausdorffness and total disconnectedness do not imply discreteness.

True. For nonempty connected spaces $X,Y$, choose $x_0\in X$, $y_0\in Y$. For each $x\in X$, the subspace $(X\times \{y_0\})\cup (\{x\}\times Y)$ is connected: its two connected pieces meet at $(x,y_0)$. The union of these subspaces over $x$ is all of $X\times Y$, and every piece contains the common connected subset $X\times \{y_0\}$. Their union is therefore connected. The assertion is also true if a factor is empty, since the empty product space is connected. This proves product of connected spaces.

True. Fix a point $x_0$ of a nonempty connected open set and let $P$ be its path-connected component. Each point of the open set has a small Euclidean ball contained in it. A ball is convex and path-connected, so each such ball lies entirely in one path component. All path components are therefore open, and the complement of $P$ is open as well. If the complement were nonempty, these two open sets would disconnect the original set. Thus $P$ is the whole set. In fact the joining paths may be chosen polygonal.

The Poynting vector in the annular region isIt points from the battery towards the resistor with the preceding current convention. Integrating through an annular cross-section givesThe energy is transported by the electromagnetic field in the space between the ideal conductors; the integration recovers the power dissipated at the terminating resistor.

Take the inner conductor to be at potential $V$ above the outer. A cylindrical Gauss law surface of length $\ell$ encloses charge $\lambda \ell$ and gives $E_{r}2\pi r\ell =\lambda \ell /{ϵ}_0$. The potential difference isThus the coaxial cylindrical capacitor field and charge areThe perfectly conducting surfaces have no tangential electric field in this idealized steady model.